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a)\(\dfrac{1}{6}x+\dfrac{1}{10}x-\dfrac{4}{15}x+1=0\)
\(\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right).x+1=0\)
\(\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{8}{30}\right).x+1=0\)
\(0.x+1=0\)
\(0.x=-1\)
=> Không có giá trị nào của x.
Vậy...
b)\(\left(\dfrac{1}{7}x-\dfrac{2}{7}\right).\left(-\dfrac{1}{5}x+\dfrac{3}{5}\right).\left(\dfrac{1}{3}x+\dfrac{4}{3}\right)=0\)
=> \(\dfrac{1}{7}x-\dfrac{2}{7}=0hoặc-\dfrac{1}{5}x+\dfrac{3}{5}=hoăc\dfrac{1}{3}x+\dfrac{4}{3}=0\)
+)\(~\dfrac{1}{7}x-\dfrac{2}{7}=0\) +) \(-\dfrac{1}{5}x+\dfrac{3}{5}=0\) +) \(\dfrac{1}{3}x+\dfrac{4}{3}=0\)
\(\dfrac{1}{7}x=-\dfrac{2}{7}\) \(-\dfrac{1}{5}x=-\dfrac{3}{5}\) \(\dfrac{1}{3}x=-\dfrac{4}{3}\)
\(x=2\) \(x=3\) \(x=-4\)
Vậy...
a 1/6x+1/10x-4/15x+1=0
(1/6+1/10-4/15)x+1=0
0x+1=0
0x=-1
x=-1/0
Vậy không có x (vì không có số nào chia cho 0)
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(\frac{-1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\hept{\begin{cases}\frac{1}{7}x-\frac{2}{7}=0\\\frac{-1}{5}x+\frac{3}{5}=0\\\frac{1}{3}x+\frac{4}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=3\\x=-4\end{cases}}}\)
KL
b, \(\left|\frac{5}{3}x\right|=\left|\frac{-1}{6}\right|\)
\(\left|\frac{5}{3}x\right|=\frac{1}{6}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}x=\frac{1}{6}\\\frac{5}{3}x=\frac{-1}{6}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}\\x=\frac{-1}{10}\end{cases}}}\)
KL
c, \(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\left|\frac{-3}{4}\right|\)
\(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\frac{3}{4}\)
\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|=\frac{3}{2}\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\\frac{3}{4}x-\frac{3}{4}=\frac{-3}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{-3}{4}\end{cases}}}\)
KL
Căng, sự thật là nó rất căng
Nhg dù sao thì.....
1) \(A\left(x\right)=\left(x-4\right)^2-\left(2x+1\right)^2\)
Xét \(A\left(x\right)=0\)
\(\Rightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Rightarrow x^2-8x+16-4x^2-4x-1=0\)
\(\Rightarrow-3x^2-12x+15=0\)
\(\Rightarrow-3x^2+3x-15x+15=0\)
\(\Rightarrow-3x\left(x-1\right)-15\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(-3x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-3x-15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
2)(Sửa đề nha, sai cmnr) \(B\left(x\right)=x^3+x^2-4x-4\)
Xét \(B\left(x\right)=0\)
\(\Rightarrow x^3+x^2-4x-4=0\)
\(\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-1\end{matrix}\right.\)
Đó là những j mình biết 
1, \(\left(x-4\right)^2-\left(2x+1\right)^2=\left(x-4-2x-1\right)\left(x-4+2x+1\right)=-3\left(x+5\right)\left(x-1\right).\)
\(\orbr{\begin{cases}x+5=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=1\end{cases}}}\)(mấy cái này áp dụng hàng đẳng thức lớp 8 mới hok)
2,\(x^3+x^2-4x-4=\left(x-2\right)\left(x^2+3x+2\right)=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
\(\orbr{\begin{cases}x=\mp2\\\end{cases}}x=-1\)
tương tụ lm tiếp nhe buồn ngủ quá rồi !
